# Drag, terminal velocity, and force.

In summary, both mice and rats have a drag coefficient of CD ≈ 0.3, meaning that they will reach terminal velocity before hitting the ground. However, the mouse will likely suffer broken legs due to a force of 1.5 × 108 N/m2, while the rat will not suffer any broken bones.

## Homework Statement

Coal miners often find mice in deep mines but rarely find rats; let’s see if we can figure out why. A mouse is roughly 5 cm long by 2 cm wide and has a mass of 30 g; a rat is roughly 20 cm long by 5 cm wide and has a mass of 500 g. Assume that both have a drag coefficient CD ≈ 0.3.

1. a) Estimate the terminal falling speeds reached by a mouse and a rat, respectively.
2. b) Assume that mine shafts are deep enough that both the mouse and the rat reach terminal velocity before hitting the bottom. Estimate the magnitude of the maximum force required to stop both the rat and the mouse when they hit the bottom. (Hint: Over what distance will the center of mass travel between the beginning of the collision and when the animal is at rest? What is the acceleration required to bring the animal to rest over this distance? The center of mass can’t travel more than about half the height of an animal, why?)
3. c) Bones will break if they are subjected to a compressional force per unit area of more than about 1.5 × 108 N/m2. A mouse may have leg bones about 1.5 mm in diameter; for a rat, they might be about twice as thick. Using your estimates from part b), determine if either the mouse or the rat (or both) will suffer broken legs. Is your answer consistent with the observations of the coal miners? Explain.

## Homework Equations

FD = ρ*v^2*CD*A/2
F = m*a
xcm = (m1*r1 + m2*r2 + ... + mn*rn)/(m1 + m2 + ... + mn)

## The Attempt at a Solution

I did part A without a problem. Simply plugged in values into the FD equation and set it to equal mg for each: mouse and rat. Mouse terminal velocity: 44.27 m/s, Rat terminal velocity: 57.155 m/s.

Part B, I'm not too sure about it. I have no idea what the hints really mean. I actually have no clue how to even start it.

Part C, using Part B's answer, I feel like I can figure it out, however, I'm not entirely sure what area to use, as it states that the leg bone is 1.5 mm in diameter, but says nothing about length, so I'm not sure what to use for surface area.

• b) Assume that mine shafts are deep enough that both the mouse and the rat reach terminal velocity before hitting the bottom. Estimate the magnitude of the maximum force required to stop both the rat and the mouse when they hit the bottom. (Hint: Over what distance will the center of mass travel between the beginning of the collision and when the animal is at rest? What is the acceleration required to bring the animal to rest over this distance? The center of mass can’t travel more than about half the height of an animal, why?)

I have no idea what the hints really mean.
mouse is roughly 5 cm long by 2 cm wide
Read the information you are given.

Bystander said:
Read the information you are given.

In class, the only center of mass we covered was between two objects, and that the center of mass never travels. Did I miss something?

Think in terms of "center of mouse."

Bystander said:
Think in terms of "center of mouse."
I'm going to have to google this. Why would the center of the mouse move between collision + at rest? Do you have any resources I can look at?

I'm going to have to google this. Why would the center of the mouse move between collision + at rest? Do you have any resources I can look at?
Read the last sentence in part B. Seems you have to assume the rodents land end-on, since you are given the lengths, not the heights. This is a bit inconsistent: to calculate the drag you had to assume they fall feet first, i.e. in their normal orientation. How, in the dark, they manage to figure out when they're about to hit the ground I'm not sure.
The other doubtful thing about the hint is that by the time the part of the animal that had been the mass centre hits ground level it would be in a fairly bad way. In practice, it would be a question of absorbing the shock over the distance by which they can extend/draw up their legs, but we have no info on this.

haruspex said:
assume the rodents land end-on, since you are given the lengths, not the heights.
My impression was that one could/should "assume a cylindrical mouse/rat" for the impact.

In class, the only center of mass we covered was between two objects, and that the center of mass never travels. Did I miss something?
A single object also has a center of mass (How could it not?).

Take a ball. Or a brick.

A ball (or brick) has mass and a center of mass. If you drop the ball (or brick), the center of mass moves with the ball (or brick). (How could it not?)

Bystander said:
My impression was that one could/should "assume a cylindrical mouse/rat" for the impact.
Maybe. Is there some particular wording that suggests that to you?

## What is drag?

Drag is a force that opposes the motion of an object through a fluid (such as air or water). It is caused by the fluid particles colliding with the surface of the object, creating a resistance to its movement.

## What is terminal velocity?

Terminal velocity is the maximum velocity that an object can reach when falling through a fluid. It occurs when the drag force is equal to the force of gravity, resulting in a constant velocity.

## What factors affect drag?

The factors that affect drag include the shape and size of the object, the density and viscosity of the fluid, and the speed of the object through the fluid. Objects with larger surface areas, higher speeds, and denser fluids tend to experience higher levels of drag.

## How does drag affect an object's motion?

Drag can either slow down or speed up an object's motion, depending on the direction of the force. If drag is in the same direction as the object's motion, it will cause it to slow down. If drag is in the opposite direction, it can aid in speeding up the object.

## How is drag calculated?

The drag force on an object can be calculated using the drag equation, which takes into account the object's shape, size, speed, and the properties of the fluid it is moving through. The equation is Fd = 1/2 * ρ * v^2 * Cd * A, where ρ is the fluid density, v is the velocity, Cd is the drag coefficient, and A is the object's cross-sectional area.

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